/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 10 The joint probability density fu... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 6: Problem 10

The joint probability density function of \(X\) and \(Y\) is given b $$ f(x, y)=e^{-(x+y)} \quad 0 \leq x<\infty, 0 \leq y<\infty $$ Find (a) \(P\\{X

Short Answer

Expert verified
The short answer is as follows: (a) The probability \(P\{X<Y\} = \frac{1}{2}\). (b) The probability \(P\{X<a\} = -(e^{-a}-1)\).

Step by step solution

01

(a) Calculate \(P\{X

To calculate the probability that \(X<Y\), we need to integrate the joint pdf over the region where this condition holds true. The region of integration will be the area above the line \(x = y\) (since \(x < y\)) in the first quadrant. The probability can be found by the following double integral: \(P\{X<Y\} = \int_{0}^{\infty} \int_{x}^{\infty} e^{-(x+y)} dy\,dx\) Now let鈥檚 solve the integral: Step 1: First, solve the inner integral w.r.t. y: \(\int_{x}^{\infty} e^{-(x+y)} dy = -e^{-(x+y)} \Big|_{x}^{\infty} = e^{-2x}\) Step 2: Now, solve the outer integral w.r.t. x: \(P\{X<Y\} = \int_{0}^{\infty} e^{-2x} dx = -\frac{1}{2}e^{-2x}\Big|_0^{\infty} = -\frac{1}{2}(0-1) = \frac{1}{2}\) Thus, the probability \(P\{X<Y\} = \frac{1}{2}\).
02

(b) Calculate \(P\{X

Now, let's find the probability of \(X<a\) by integrating the joint pdf over the range where \(0 \leq x < a\) and \(0 \leq y < \infty\). To calculate the probability, we can use the following double integral: \(P\{X<a\} = \int_{0}^{a}\int_{0}^{\infty} e^{-(x+y)} dy\, dx\) Let's solve the integral: Step 1: First, solve the inner integral w.r.t. y: \(\int_0^{\infty} e^{-(x+y)} dy = -e^{-(x+y)}\Big|_{0}^{\infty} = e^{-x}\) Step 2: Now, solve the outer integral w.r.t. x: \(P\{X<a\} = \int_{0}^{a} e^{-x} dx = -e^{-x}\Big|_0^{a} = -(e^{-a}-1)\) Thus, the probability \(P\{X<a\} = -(e^{-a}-1)\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Probability Integration
Probability integration represents a fundamental concept in probability theory, which involves calculating the likelihood of a particular outcome by integrating a probability density function (pdf) over a given range. When we are dealing with continuous random variables, the probability that a variable falls within a certain interval is assessed through integration.

In the context of our exercise, we're looking at the joint probability density function (pdf) for two variables, \(X\) and \(Y\), which is performed using a double integral. Double integrals appear when we're estimating the probability of outcomes over a two-dimensional region. When you integrate such a function over a desired range, you're essentially summing up all the probabilities within that area to find the total likelihood of a certain event occurring.
Double Integral
A double integral is used in probability theory when we work with joint probability density functions of two variables. It's a mathematical tool that allows us to calculate the accumulation of a function over a two-dimensional area.

In the given exercise, to find the probability of \(X
Exponential Distribution

Characteristics of Exponential Distribution


The exponential distribution is a continuous probability distribution that is often used to model the time between events in a Poisson process. It is characterized by its simplicity and has a parameter called the rate parameter, typically denoted as \(\theta\). The probability density function of an exponential distribution is \(f(x) = \theta e^{-\theta x}\), for \(x \textgreater 0\), and \(f(x) = 0\) elsewhere.

Relevance to the Exercise


In our exercise, the joint pdf has the form \(f(x, y)=e^{-(x+y)}\), which suggests that both \(X\) and \(Y\) are exponentially distributed and independent. The exponential distribution is memoryless, which means that past events do not affect the probability of future events. This is key to solving problems such as the one in the exercise, as the independence of the variables simplifies the integration process.
Probability Theory
Probability theory is the branch of mathematics concerned with analysis of random phenomena. The central objects of probability theory are random variables, events, and stochastic processes. The theory is based on probability measures, which are models for phenomena where we need to quantify the likelihood of different outcomes.

In exercises like the one provided, we apply concepts from probability theory to solve for specific probabilities using the joint probability density function of two random variables. This theory underpins the methodology to integrate over certain regions and understand dependencies or independences between variables (\(X\) and \(Y\) in our case). The foundation laid by probability theory makes it possible to model real-world situations mathematically and extract meaningful conclusions about the behavior of complex systems.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Let \(X_{(1)} \leq X_{(2)} \leq \cdots \leq X_{(n)}\) be the ordered values of \(n\) independent uniform \((0,1)\) random variables. Prove that for \(1 \leq k \leq n+1\), $$ P\left\\{X_{(k)}-X_{(k-1)}>t\right\\}=(1-t)^{n} $$ where \(X_{0} \equiv 0, X_{n+1} \equiv t\).

Each throw of a unfair die lands on each of the odd numbers \(1,3,5\) with probability \(C_{3}\) and on each of the even numbers with probability \(2 C\). (a) Find \(C\). (b) Suppose that the die is tossed. Let \(X\) equal 1 if the result is an even -number, and let it be 0 otherwise. Also, let \(Y\) equal 1 if the result is a number greater than three and let it be 0 otherwise. Find the joint probability mass function of \(X\) and \(Y\). Suppose now that 12 independent tosses of the die are made. e(c) Find the probability that each of the six outcomes occurs exactly twice. (d) Find the probability that 4 of the outcomes are either one or two, 4 are either three or four, and 4 are either five or six. (e) Find the probability that at least 8 of the tosses land on even numbers.

A rectangular array of \(m n\) numbers arranged in \(n\) rows, each consisting of \(m\) columns, is said to contain a saddlepoint if there is a number that is both the minimum of its row and the maximum of its column. For instance, in the array $$ \begin{array}{rr} 1 & 3 & 2 \\ 0 & -2 & 6 \\ .5 & 12 & 3 \end{array} $$ the number 1 in the first row, first column is a saddlepoint. The existence of a saddlepoint is of significance in the theory of games. Consider a rectangular array of numbers as described above and suppose that there are two individuals \(-A\) and \(B\) - that are playing the following game: \(A\) is to choose one of the numbers \(1,2, \ldots, n\) and \(B\) one of the numbers \(1,2, \ldots, m\). These choices are announced simultaneously, and if \(A\) chose \(i\) and \(B\) chose \(j\), then \(A\) wins from \(B\) the amount specified by the number in the ith row, \(j\) th column of the array. Now suppose that the array contains a saddlepoint-say the number in the row \(r\) and column \(k-\) call this number \(x_{r k}\). Now if player \(A\) chooses row \(r\), then that player can guarantee herself a win at least \(x_{r k}\) (since \(x_{r k}\) is the minimum number in the row \(r\) ). On the other hand, if player \(B\) chooses column \(k\), then he can guarantee that he will lose no more than \(x_{r k}\) (since \(x_{r k}\). is the maximum number in the column \(k\) ). Hence, as \(A\) has a way of playing that guarantees her a win of \(x_{r k}\) and as \(B\) has a way of playing that guarantees he will lose no more than \(x_{r k}\), it seems reasonable to take these two strategies as being optimal and declare that the value of the game to player \(A\) is \(x_{r k}\). If the \(n m\) numbers in the rectangular array described above are independently chosen from an arbitrary continuous distribution, what is the probability that the resulting array will contain a saddlepoint?

The joint density function of \(X\) and \(Y\) is $$ f(x, y)= \begin{cases}x y & 0

The joint density function of \(X\) and \(Y\) is $$ f(x, y)= \begin{cases}x+y & 0
See all solutions

Recommended explanations on Math Textbooks

Discrete Mathematics

Read Explanation

Statistics

Read Explanation

Geometry

Read Explanation

Applied Mathematics

Read Explanation

Decision Maths

Read Explanation

Calculus

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.